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**Rebesques** Not in general. You need the extra assumption that $\displaystyle f^n,f^{n-1}$ are never zero.

The proof is easy: Choose any natural number $\displaystyle m$. Since $\displaystyle gcd(n,n-1)=1,$ there are integers $\displaystyle q, \ p$ such that $\displaystyle pn+q(n-1)=1$. So $\displaystyle m=pmn+qm(n-1)$; Thus $\displaystyle f^m=(f^n)^{mp}(f^{n-1})^{qm}$ is differentiable as a product of differentiable functions. Note that the exponents $\displaystyle mp, \ mq$ may be negative, and so the assumption $\displaystyle f^n, f^{n-1}\neq0$ is essential.